7/06 Electric and Magnetic Forces ELECTRIC AND MAGNETIC FORCES About this lab With the discovery that electric charge comes in discrete particles with inertial mass, it became of great interest and practical importance to elucidate the mechanisms of interaction among the particles. Two forces were found, electric and magnetic, the latter affecting only moving charges. The field concept was invented to to describe the spatial aspects of charge particle interactions. Mastery of electricity has enabled modern technological life in all aspects. Examples in which the motion of charged particles is directly affected by application of electric or magnetic fields produced by other charges include cathode ray tubes (e.g. monitors) and mass spectrometers (e.g. chemical analysis). There are only two fundamental equations below: The electrical force law &gt; Equation 1a FE= q E and its electric potential correlate Equation 1b 1 2 mv = q V , and 2 The magnetic force law &gt; Equation 2 F B = q x vB v where is the particle's velocity vector. All the other equations derive from these, as applied to the particular geometries of the apparatus (in simplifying approximations): Plane parallel electrodes, and parallel circular current coils in Helmholtz configuration. Since the physical arrangement deviates from the idealizations in our equations, your experimental results will deviate from the predictions from those idealizations. Nevertheless, as various ratios will show, agreement is not bad. Appendix 1 shows simulations of beam paths in constant E , B , and E x B fields. Appendices 2 and 3 give detailed derivations and procedural details. References: Physics, Cutnell &amp; Johnson, 6 th Ed., Chapters 18, 21 (Wiley 2004) 7/06 Electric and Magnetic Forces <a href="/keyword/physics-serway/" >physics serway</a> &amp; Beichner, 5 th Ed. v.2, Chapters 23, 29 (Wiley 2000) Apparatus: Evacuated glass tube with accelerating electrode and electric deflecting plates, external magnetic deflecting current coils (Helmholtz configuration), power supplies, meters, Vernier Software's Graphical Analysis CAUTION!!! THIS EXPERIMENT USES A VERY HIGH VOLTAGE POWER SUPPLY! BE EXTREMELY CAREFUL! DO NOT MAKE CONNECTIONS WITH THE POWER SUPPLY ON. DO NOT TOUCH THE HV WIRE OR ITS CONNECTORS WHEN THE SUPPLY IS ON. DON'T ALTER FILAMENTS. WHEN IN DOUBT - ASK YOUR INSTRUCTOR FIGURE 1: The fluorescent screen is gridded. The electron beam path produces fluorescence. You will read (x,y) coordinates along the path of a deflected beam (x horizontal, y vertical) and enter into a Graphical Analysis file, which will apply the path theory appropriate to the physical case and calculate quantities of interest. (In the simulations below, motion is in the (x,z) plane, whereas in our experimental axes, it is in the (x,y) plane. 7/06 There are three possible arrangements: Electric and Magnetic Forces Part A : Electric Deflection Field Only As the electrons in the deflection region move in the horizontal (x) direction at constant speed, after acceleration to constant x-speed by potential V a , an electric force may be applied in the vertical (y) direction by applying a deflecting potential difference V d between two horizontal deflecting plates, producing a changing v y (constant, non-zero Ey ), with constant v x . The electron &quot;falls&quot; upward under electric force, gaining speed, as a horizontally fired projectile falls downward under gravitational force. As for a projectile moving under constant gravitational acceleration, the path is parabolic. The path does not depend on things such as the electron charge to mass ratio deflecting voltage V d , and the accelerating voltage V a plate separation d : y= x 4d 2 e , the m , but only involves the electric Electric deflection path theory (which so the only information fitting the deflection parabola gives is the value of d can be obtained more easily and reliably by looking at the apparatus! This surprising result comes from the electrical system, in which there is only a single electrical power supply for both deflection and accelerating voltages: V d = V a . See below for a detailed derivation. Part B: Magnetic Deflection Field Only In addition, the apparatus permits deflection in the vertical plane (both x and y directions) by a magnetic force, produced by current in an external &quot;Helmholtz coil&quot; arrangement. A constant, external magnetic field acts at right angles to the motion, producing acceleration without work, so the speed remains constant. The path is circular. As with a mass whirled on a string with constant tension, the radial magnetic force on the 2 mv moving electron charge = electron mass x radial acceleration: evB = , giving R R= mv where, assuming the circle point is (0,0), and any path (x,y) point gives R: eB 2 2 x +y R= ( ). 2y 7/06 See below for a detailed derivation. Electric and Magnetic Forces Part C: Crossed Electric and Magnetic Deflection Fields And, electric and magnetic deflections may be applied simultaneously. In general, the E and B forces are unbalanced, and the path is curved. If there is balance between E and B forces, the Wien velocity filter path is a straight line: F E = - F B or e | E | = ev | B | , with the correct relative directions of E and B . See below for details. Appendix 1 Figure 2 Simulation, Case A. Electric field only. Quadratic beam path under constant acceleration. The particle deflects opposite to to the E field direction, because it is negatively charged. The path is parabolic (quadratic fit). 7/06 Electric and Magnetic Forces Figure 3 Simulation, Case B Constant external magnetic field only. The particle moves in a circle perpendicular to the B field, at constant speed but non-constant velocity. The B field accelerates the particle (changes its direction) but does no work on it, because the velocity is always perpendicular to the field. Note negative charge. The path is circular. In the actual experiment, only a small part of the circular path will be observed. 7/06 Electric and Magnetic Forces Figure 4a Simulation, Case C (Wien velocity filter) Magnetic and electric forces are balanced, producing no deflection for the particular |E | particle velocity. Check ratio and compare with velocity. Note negative |B| charge. The path is linear. 7/06 Electric and Magnetic Forces Figure 4b Simulation: Case C ExB fields. B field too large to produce straight |E | path, for the particle's velocity. Check ratio and compare with velocity. |B| Note negative charge. http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=53 7/06 Electric and Magnetic Forces Introduction: The experiment employs a vacuum tube which allows unimpeded motion of electrons under the influence of applied electromagnetic fields. The tube incorporates an &quot;electron gun&quot;. An oxidized metal filament is heated (note glow in tube neck at right) so that some of the electrons of the metal get enough energy to leave the filament. They are then accelerated by an accelerating potential difference Va, which gives them kinetic energy mvx2, equal to eVa . (e is charge of an electron, -1.6x10-19 coulombs, and m is its mass, 9.1 x 10-31 kg.) The electron beam is made visible when the electrons strike a fluorescent screen. A voltage of one to three thousand volts is applied across two horizontal conducting plates which are insulated from each other. The upper plate is a=, the lower (ground). Current coils in series provide a horizontal magnetic field in the deflection region. Figure 5 Evacuated tube for electron acceleration and evacuation, with electrical connections and external Helmholtz magnetic field current coils. Not shown is the internal hot filament electron source, which produces the glow seen innoperation. Copy the supplied Graphical Analysis file and save under another name. Enter data in each part as directed. Observe the graphical presentation of results and the ratios which give the relation between your experimental results, as interpreted in the simplifying geometric approximations to the actual physical system, and the results of assuming the simplifying results are correct. These ratios are typically around 1. Appendix 2 Detailed derivations 7/06 Electric and Magnetic Forces Part A Electric Deflection Field Only Simplified model A: Constant field with sharp boundary at x = 0. Electrons enter electric field region with vertical E full acceleration potential energy. We will approximate the finite parallel conducting plate geometry using the infinite plate V result that E = d is constant and perpendicular to the plates. Then there is no d horizontal x acceleration of a charged particle, and constant vertical y acceleration (upward toward + plate for charge). Deflection Force Equation 1a: F = q E = m a theoretical --&gt; a th = constant, in y direction Then a th = F E e Vd . =q = m m md 1 2 y = a th t for horizontal entry to field region, with 2 2 mx x 2 1 e Vd 2 y= t and t = 2 = v x 2eV a 2m d 2 from the 1 2 mv x = eV a 2 Acceleration Energy Conservation: Equation 1b: So, the electron path is quadratic in time, and also in x because x is proportional to time t: y= 2 2 1 e V d mx 1 1 Vd 2 x = x= 2 m d 2eV a 2 2d V a 4d because the circuits are wired so that V d = V a . Note that,for this reason, the path is independent of accelerating voltage V a . If there Vd were two independent power supplies, the ratio could be varied, and so the path. Va Why does not this expression contain the electron charge or mass? Because the e horizontal acceleration voltage V A acts on the same as does the vertical m acceleration voltage V d . Why do the voltages V a and V d not appear? To repeat: because a single high voltage power supply is used for both, for simplicity and Vd economy. Otherwise the ratio would appear and the path would generally be Va 7/06 Electric and Magnetic Forces different than ours. But, as long as the voltage ratio did not change, neither would the trace. Turn up V a ( = V d ) until you see a clear blue beam trace. Read several pair values ( x , y ) on the trace (in meters), and enter into Graphical Analysis Part A. Graph y vs. x; Analyze:Curve Fit: Quadratic. (GA's quadratic fit function has the form a GA x 2 + bx + c where a GA is just a generic coefficient of x 2 . Compare a GA by ratio with 1 : 4d 1 / a GA . 4d The deflection plates are not infinite and, although they are mechanically placed symmetrically placed above and below the system axis, they are not electrically symmetric the upper plate is at acceleration potential V a , the same as the acceleration and beam defining structure, and the lower plate is at ground potential 0 (power supply reference) voltage. Part B Magnetic Deflection Field Only Simplified Model B: Constant horizontal B field with sharp cutoff at x = 0. Electrons enter magnetic field region with full acceleration potential energy. A static magnetic field exerts a velocity dependent force on a moving charge at right angles to the motion, so it changes the direction of motion (i.e., the velocity direction) without doing any work on the particle (so constant velocity magnitude), so circular motion results: Magnetic Force: Equation 2 v | F | = evB = m R 2 --&gt; R = mv eB We get velocity v from acceleration Equation 2 and | B | from coils in series. | B | = 4.23x10 3 1 mv 2 = eV a 2 --&gt; v = 2eV a m I where I is the common current through the two If we could see an entire circular path, we could determine R and thus an e exp to compare with the accepted value. But we can't. experimental value of m However, a partial arc of the circle should do. In fact, two points on the circle should do, along with the assumption that the circle's center lies directly above the x = 0 entrance point. (Remember, we assumed sharp field cut-offs at that point, for simplicity. See the figure in Appendix 2.) We will follow this approach (i.e., two points of the circle), finding 7/06 Electric and Magnetic Forces different bending radii R for a few different | B | values (different currents I ). We will also reverse coil current direction to partially cancel effects of coil misalignment and of (small) earth magnetic field components. If we could measure x and y (points on the circle) from the center, x and y would be related by x 2 + y 2 = R2 But our x , y origin ( 0,0 ) lies at the bottom of the circle, so the relation is (after a little geometry) R= x y 2y 2 2 . ( 0,0 ) will be one of our two points for each circle, and various observed ( x , y ) values along the circle will be the other. (These should give the same R , in principle.) Then 2V e = 2 a2 m B R where R =( 2 x +y 2 ) and | B | = 4.23x10 3 I . 2y 2 2 Set and record fixed V a at fixed I B ; then observe and enter ( x , y ) and I B for four well-spaced values of x over the circular beam path. Substitute your value of e exp Reverse the current the acceleration voltage V a into the definition of m direction and repeat ( y signs should reverse) . (The + and current values should have approximately equal magnitudes, to average asymmetries and earth field effects.) GA will calculate R's for each entry, but you must insert your own operating value of e exp . accelerating voltage V a into the definition of calculated column m e exp vs. observation number N. Examine for gross m e exp and the standard discrepancies. Analyze:Statistics. Record the mean value of m deviation. GA will also plot the ratio GA will calculate and plot the ratio e e e 11 exp / accepted = exp / 1.76x10 . m m m Observe for systematic trend. Analyze: Statistics. Is the mean ratio near to 1? 7/06 Electric and Magnetic Forces Note: There is no theory giving numerical values of e and m; they are taken as fundamental experimental facts of our universe. String theory is trying. Part C Crossed Electric and Magnetic Deflection Fields Simplified model C: field and constant horizontal B , each with sharp boundary at x = Constant vertical E 0. Electrons enter electric field region with full acceleration potential energy. The electric and magnetic fields are at right angles. But the magnetic field force produces force at right angles to its direction (and that of the velocity vector), so both fields act in the vertical direction, either aiding or opposing. We will seek the current direction so that the two forces oppose and so that the two force magnitudes cancel, producing zero deflection. From the electric and magnetic force equations |E | , a velocity condition independent of charge magnitude or |B| sign. Such a condition is obtained (by adjustment of E , B or both to select beam particles of a desired velocity (Wien velocity filter). Others will be deflected one way or the other, according as their velocities are greater or less than that selected for. q E | = qv | | or | B v= Set a few accelerating voltages V a , then adjust magnetic field B in each case by adjustment of coil current I to obtain zero deflection (this won't be exact -look for horizontal beam in center of deflection region). Then Vd Va |E | 3 3 v Wien = = / 4.23x10 I = / 4.23x10 I (remember V d = V a ), which |B| d d can be compared with the velocity predicted from the acceleration energy conservation Equation 2 above va = e 2Va . m Enter into Graphical Analysis Part C several values of V a and I that produce zero deflection, and let GA calculate the Wien velocity v Wien and the comparison velocity v a , obtained from acceleration energy. In calculating v a , we will use the accepted e 11 = 1.76x10 mks units . value m accepted 7/06 Electric and Magnetic Forces v Wien . Examine the plot for any systematic va trend. Analyze: Statistics. Does the mean ratio differ substantially from 1? GA will also calculate and plot the ratio Appendix 3 Detailed procedures A. ELECTRIC DEFLECTION ONLY Figure 6 Part A quadratic fit to electric field beam deflection. 7/06 Electric and Magnetic Forces Figure 7 Part A (Electric deflection) - top deflection plate is connected to the positive terminal of the large high voltage power supply (white cable), which also provides the acceleration voltage Va (Va = Vd). The bottom deflection plate is connected to the negative terminal (ground) of the large power supply (black cable). Turn all voltages to zero and turn off the power supply. Turn off the magnet power supply. With the high voltage power supply turned down and off, connect the + high voltage cable (white) from the high voltage power supply to the top deflection plate electrode feed through Connect the ground (-) cable (black) from the power supply to the bottom plate. Note again that the same high voltage is applied both to the electron gun and to the deflection plates. Do not turn the power supply on until the instructor has checked your connections. Check that both the high voltage and the coil current control knobs are turned fully counter-clockwise. Turn on the power supply and observe the beam trace while varying the high voltage. At Busch, the small multimeter should be set to a scale of 20 V and its reading (in volts) should be multiplied by 500. At Douglass, use the 200 mV scale and multiply the meter reading in millivolts by 100 to get the high voltage output. (At Busch, a meter reading of 4 V --&gt; 2,000 volts; at Douglass, a meter reading of 20 mV --&gt; 2,000 volts.) Verify that the beam trace locus does not depend on accelerating voltage. Set Va to the lowest voltage that gives a clear and stable trace and leave it fixed. Enter into GA the 7/06 Electric and Magnetic Forces the values of y beam at x beam = 5, 6, . . . . cm. Remember to convert everything to SI units (specifically meters); mixing units will give you an erroneous result. Why does the beam trace intensity increase with increasing high voltage? B. MAGNETIC FIELD DEFLECTION ONLY Part B (Magnetic deflection) See Figure 7. Turn high voltage down and off. Disconnect the black ground wire!!! Connect bottom deflection plate to top deflection plate via jumper cable (short cable with alligator clips at both ends). Helmholtz Coils A useful laboratory technique for getting a fairly uniform magnetic field is to use a pair of circular coils on a common ...